How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola?

Question

How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !

Answer 1

Taking the '+' sign, $$y^2-x^2=x+y+1\implies\left(y-\frac12\right)^2-\left(x+\frac12\right)^2=1^2$$

$$\implies\frac{\left(y-\frac12\right)^2}{1^2}-\frac{\left(x+\frac12\right)^2}{1^2}=1^2$$

Similarly for the '-' sign,

$$y^2-x^2=-(x+y+1)\implies\left(x-\frac12\right)^2-\left(y+\frac12\right)^2=1^2$$

Can you recognize $a,b$ here?

Answer 2

Let $$ \frac{y^2-x^2}{x+y+1}=1\\ \Rightarrow y^2-x^2=x+y+1\\ \Rightarrow y^2-x^2-x-y=1 $$ Complete the squares for x and y . You will get rectangular hyperbola. Similar will be the case if $$ \frac{y^2-x^2}{x+y+1}=-1$$